Let say you need to calculate the reinforcement for a circular slab of 2 Metre diameter with spacing 200 mm c/c and clear cover as 25 mm.

Unlike other RCC members, this one is different since the rod length will differ according to the diameter of the circle.

We need to calculate the length of the required steel using Pythagoras theorem slightly different from normal one like painting area coefficient.

## Circular Slab Reinforcement Detailing

Here, Where L1 = Diameter of the Circular Slab Reinforcement

To find L2, we have to use Pythagoras theorem,

L2 = √(R^{2} – h_{1}^{2}) X 2

The same goes to L3 = √(R2 – h_{2}^{2}) X 2

L4 = √(R^{2} – h_{3}^{2}) X 2

And so on..

Where L is the length of the reinforcement bar,

R – Radius of the circular slab reinforcement ( Excluding clear cover on both sides)

H – Distance between the bar to the main bar

Note – We need to calculate the Typical L’s for the 1st half only, for the 2nd half we will multiply it by 2 excluding the main bar.

# Circular Slab Reinforcement Calculation

Assume We need to calculate the reinforcement for a circular slab of 2 Metre diameter with 12 mm main rod and 10 mm distribution rod @ spacing 200 mm and clear cover as 25 mm.

First of all, you have to calculate the number of rods that means, how many L’s we need to calculate.

That’s Simple,

Take the diameter of the circular reinforcement = Diameter of circular slab – Clear Cover on Both Sides

= 2000 mm – 25 mm – 25 mm = 1950 mm

We know, L = 1950 mm or 1.95 metre

R = L/2 = 1.95 m /2 = 0.975 m

Number of L’s = Radius of the circular reinforcement / Centre to centre distance = 0.975 / 0.200 = 4.87 No’s

Number of rod = 5 – centre rod (since centre rod will be only one)

= 4 X 2 (2nd half) = 8 No’s

Now we need to calculate 8 number of main rods and 8 numbers of distribution rods. Refer the picture

As we know, L1 = Diameter of the main bar = 1.95 m

the formula for L2 = √(R^{2} – h_{1}^{2}) X 2 = √(0.975)2 – (0.2)2 X 2 = 1.87 m

L3 = √(R^{2} – h_{2}^{2}) X 2 = √(0.975)2 – (0.4)2 X 2 = 1.63 m

L4 = √(R^{2} – h_{3}^{2}) X 2 = √(0.975)2 – (0.6)2 X 2 = 1.23 m

L5 = √(R^{2} – h_{4}^{2}) X 2 = √(0.975)2 – (0.8)2 X 2 = 0.67 m

Reinforcement for 1st half L = L2+L3+L4 = 1.87 m + 1.63 m +1.23 m +0.67 m = 5.4 m

Reinforcement for 2nd half = ( 1st Half X 2 ) = 5.4 m X 2 = 10.8 m

Total Main rod Reinforcement = 1st half + 2nd half + Centre Rod (L1) = 10.8 m + 1.95 m = 12.75 m

So now we have the reinforcement details for the main distribution.

Again we need to multiply it by 2 for the distribution rods.

## Bar Bending Schedule for Circular Slab

BAR | NUMBERS | LENGTH | DIAMETER | REINFORCEMENT |
---|---|---|---|---|

Main Bar |
||||

L1 | 1 | 1.95 m | 12 mm | 1.95 m |

L2 | 2 | 1.87 m | 12 mm | 3.74 m |

L3 | 2 | 1.63 m | 12 mm | 3.26 m |

L4 | 2 | 1.23 m | 12 mm | 2.46 m |

L5 | 2 | 0.67 m | 12 mm | 1.34 m |

Distribution Bar |
||||

L1 | 1 | 1.95 m | 10 mm | 1.95 m |

L2 | 2 | 1.87 m | 10 mm | 3.74 m |

L3 | 2 | 1.63 m | 10 mm | 3.26 m |

L4 | 2 | 1.23 m | 10 mm | 2.46 m |

L5 | 2 | 0.67 m | 10 mm | 1.34 m |

We have explained how to calculate a weight of a steel bar?

**Note** – If you are not sure about how many rods by dividing, keep on doing L formula if you have negative value then that is the end of your calculation.

Hope you have enjoyed the content.

Share with your friends. Happy Learning 🙂